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Question Number 47058 by maxmathsup by imad last updated on 04/Nov/18

$$calculate{\int }_0^1\frac{1+x^3}{1+x^4}dx$$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18

$$\int \frac{dx}{1+x^4}+\frac{1}{4}\int \frac{4x^3}{1+x^4}dx$$$$I_1+I_2$$$$I_2=\frac{1}{4}\int \frac{4x^3}{1+x^4}dx$$$$=\frac{1}{4}ln(1+x^4)+c_2$$$$j_2=\mid I_2{\mid }_0^1=\frac{1}{4}ln(1+1)-\frac{1}{4}ln(1+0)\to \frac{1}{4}ln2$$$$I_1=\frac{1}{2}\int \frac{\frac{2}{x^2}}{\frac{1}{x^2}+x^2}dx$$$$=\frac{1}{2}[\int \frac{(1+\frac{1}{x^2})-(1-\frac{1}{x^2})}{x^2+\frac{1}{x^2}}dx$$$$=\frac{1}{2}[\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2}-\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-2}]$$$$=\frac{1}{2}[\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)-\frac{1}{2\sqrt{2}}ln\mid \left(\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right)\mid ]+c_1$$$$=\frac{1}{2}[\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{x^2-1}{x\sqrt{2}}\right)-\frac{1}{2\sqrt{2}}ln\mid \left(\frac{x^2+1-x\sqrt{2}}{x^2+1+x\sqrt{2}}\right)\mid$$$$j_1=\mid I_1{\mid }_0^1$$$$j_1=\frac{1}{2}[\{\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{1-1}{1\times \sqrt{2}}\right)-\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{0-1}{0\times \sqrt{2}}\right)\}-$$$$\frac{1}{2\sqrt{2}}ln\mid \left(\frac{1+1-\sqrt{2}}{1+1+\sqrt{2}}\right)\mid -\frac{1}{2\sqrt{2}}ln\mid \left(\frac{1}{1}\right)]$$$$=\frac{1}{2}[\{\frac{1}{\sqrt{2}}\times 0-\frac{1}{\sqrt{2}}(-\frac{\pi }{2})\}-\frac{1}{2\sqrt{2}}ln\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)]$$$$=\frac{\pi }{4\sqrt{2}}-\frac{1}{2\sqrt{2}}ln\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)$$$$$$$$nowputlimit...$$

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